A teacher shouting at the top of his lungs emits about 1.0 W of energy as sound waves. Student A is standing 2.0 m away. Student B is standing 20 m away. Compare the sound intensity level at these two locations. OB1 =B₂ B₁ = 10B2 none of the given 10ß1 = ³2 100ß1 = ³2 B1 = 10032
A teacher shouting at the top of his lungs emits about 1.0 W of energy as sound waves. Student A is standing 2.0 m away. Student B is standing 20 m away. Compare the sound intensity level at these two locations. OB1 =B₂ B₁ = 10B2 none of the given 10ß1 = ³2 100ß1 = ³2 B1 = 10032
Physics for Scientists and Engineers
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ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter16: Wave Motion
Section: Chapter Questions
Problem 54AP: Consider the following wave function in SI units: P(r,t)=(25.0r)sin(1.36r2030t) Explain how this...
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A seiche is an oscillating standing wave in a body of water. The term seiche pronounced saysh) can be understood by the sloshing of water back and forth in a swimming pool. The same phenomenon happens on a much larger scale in vast bodies of water including bays and lakes. A seizure can happen in any enclosed or semi-enclosed body of water.
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![**Sound Intensity Level Comparison**
---
**Question:**
A teacher shouting at the top of his lungs emits about 1.0 W of energy as sound waves. Student A is standing 2.0 m away. Student B is standing 20 m away. Compare the sound intensity level at these two locations.
---
**Answer Choices:**
- ○ β₁ = β₂
- ○ β₁ = 10β₂
- ○ none of the given
- ○ 10β₁ = β₂
- ○ 100β₁ = β₂
- ○ β₁ = 100β₂
---
**Explanation:**
To solve this question, we need to understand that the intensity of sound decreases with distance. The intensity (I) of a sound is inversely proportional to the square of the distance (r) from the source. This relationship can be given by the formula:
\[ I \propto \frac{1}{r^2} \]
In this context, β₁ represents the sound intensity level at Student A's location (2.0 m away), and β₂ represents the sound intensity level at Student B's location (20 m away). Because the energy of sound dissipates as it moves further away from the source, the sound intensity levels at these two points are related as follows:
- For Student A (r₁ = 2.0 m), the intensity I₁ is proportional to \(\frac{1}{2^2} = \frac{1}{4}\)
- For Student B (r₂ = 20 m), the intensity I₂ is proportional to \(\frac{1}{20^2} = \frac{1}{400}\)
Thus, you can see the ratio of the intensities can be calculated:
\[ \frac{I₁}{I₂} = \frac{\frac{1}{4}}{\frac{1}{400}} = 100 \]
This means that the sound intensity level at Student A's location (2.0 m away) is 100 times that at Student B's location (20 m away).
Thus, the correct answer according to our options is:
- ○ β₁ = 100β₂](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6db204cf-6d84-4f46-99cd-7f54c3524774%2F62e9c427-4d26-49d2-bc6f-1db26b490162%2Fcclhic_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Sound Intensity Level Comparison**
---
**Question:**
A teacher shouting at the top of his lungs emits about 1.0 W of energy as sound waves. Student A is standing 2.0 m away. Student B is standing 20 m away. Compare the sound intensity level at these two locations.
---
**Answer Choices:**
- ○ β₁ = β₂
- ○ β₁ = 10β₂
- ○ none of the given
- ○ 10β₁ = β₂
- ○ 100β₁ = β₂
- ○ β₁ = 100β₂
---
**Explanation:**
To solve this question, we need to understand that the intensity of sound decreases with distance. The intensity (I) of a sound is inversely proportional to the square of the distance (r) from the source. This relationship can be given by the formula:
\[ I \propto \frac{1}{r^2} \]
In this context, β₁ represents the sound intensity level at Student A's location (2.0 m away), and β₂ represents the sound intensity level at Student B's location (20 m away). Because the energy of sound dissipates as it moves further away from the source, the sound intensity levels at these two points are related as follows:
- For Student A (r₁ = 2.0 m), the intensity I₁ is proportional to \(\frac{1}{2^2} = \frac{1}{4}\)
- For Student B (r₂ = 20 m), the intensity I₂ is proportional to \(\frac{1}{20^2} = \frac{1}{400}\)
Thus, you can see the ratio of the intensities can be calculated:
\[ \frac{I₁}{I₂} = \frac{\frac{1}{4}}{\frac{1}{400}} = 100 \]
This means that the sound intensity level at Student A's location (2.0 m away) is 100 times that at Student B's location (20 m away).
Thus, the correct answer according to our options is:
- ○ β₁ = 100β₂
![### Sound Intensity Level Conversion Problem
You are overhearing a very heated conversation that registers 80 dB. You walk some distance away so that the intensity decreases by a factor of 100. What is the sound intensity level now?
- [ ] 20 dB
- [ ] 100 dB
- [ ] 90 dB
- [ ] 70 dB
- [ ] 60 dB
- [ ] 0 dB
- [ ] 50 dB
- [ ] 30 dB
- [ ] 80 dB
#### Explanation:
The sound intensity level, measured in decibels (dB), is a logarithmic measure of the intensity of a sound. The formula for sound intensity level is given by:
\[ L = 10 \log \left( \frac{I}{I_0} \right) \]
where:
- \( L \) is the sound intensity level in decibels (dB),
- \( I \) is the sound intensity,
- \( I_0 \) is the reference sound intensity, usually the threshold of hearing.
Given:
- Initial sound intensity level \( L_i = 80 \text{ dB} \)
- Intensity decreases by a factor of 100.
To find the new sound intensity level \( L_f \), we use the relationship of the logarithmic scale:
\[ L_f = L_i - 10 \log(100) \]
\[ L_f = 80 - 10 \cdot 2 \]
\[ L_f = 80 - 20 \]
\[ L_f = 60 \text{ dB} \]
So, the new sound intensity level is:
- [x] 60 dB](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6db204cf-6d84-4f46-99cd-7f54c3524774%2F62e9c427-4d26-49d2-bc6f-1db26b490162%2F4201yvxm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Sound Intensity Level Conversion Problem
You are overhearing a very heated conversation that registers 80 dB. You walk some distance away so that the intensity decreases by a factor of 100. What is the sound intensity level now?
- [ ] 20 dB
- [ ] 100 dB
- [ ] 90 dB
- [ ] 70 dB
- [ ] 60 dB
- [ ] 0 dB
- [ ] 50 dB
- [ ] 30 dB
- [ ] 80 dB
#### Explanation:
The sound intensity level, measured in decibels (dB), is a logarithmic measure of the intensity of a sound. The formula for sound intensity level is given by:
\[ L = 10 \log \left( \frac{I}{I_0} \right) \]
where:
- \( L \) is the sound intensity level in decibels (dB),
- \( I \) is the sound intensity,
- \( I_0 \) is the reference sound intensity, usually the threshold of hearing.
Given:
- Initial sound intensity level \( L_i = 80 \text{ dB} \)
- Intensity decreases by a factor of 100.
To find the new sound intensity level \( L_f \), we use the relationship of the logarithmic scale:
\[ L_f = L_i - 10 \log(100) \]
\[ L_f = 80 - 10 \cdot 2 \]
\[ L_f = 80 - 20 \]
\[ L_f = 60 \text{ dB} \]
So, the new sound intensity level is:
- [x] 60 dB
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